Angular momentum of skaters

1. The problem statement, all variables and given/known data
Eight 60-kg skaters join hands and skate down an ice rink at 4.6 m/s. Side by side, they form a line 12m long. The skater at one end stops abruptly, and the line proceeds to rotate rigidly about that skater. Find the angular speed.

3. The attempt at a solution
I said that initial angular momentum = [tex]rmv = 6m \times 480kg \times 4.6m/s = 13248 kg m^2 /s[/tex]
final angular momentum = [tex]\frac{1}{3} \times 480 \times 12^2 \times \omega=23040\omega[/tex]
so
[tex]13248=23040\omega[/tex]
[tex]\omega=0.575 rad / sec[/tex]
which according to the book is wrong. In the book it said 0.537 rad /sec. What have I done wrong here?

moment of inertia through the centre of a rod is
[tex]L=\frac{1}{12}ML^2[/tex]
i suppose you could pop off a 60kg here and there to get mass= 420kg but that doesnt make much sense to me because of my initial guesses. if it truly is a rod rotating right at the end of it then i thought i had to use
[tex]L=\frac{1}{3}ML^2[/tex]
as for the initial momentum i couldnt make another guess of what it could be